Metric Spaces and Compactness

This is a review of part of chapter 2 of Rudin’s introductory book on principle of mathematical analysis.

Metric

A metric on a set $X$ is a function $d : X \times X \rightarrow \mathbb{R}$ such that

$d(x,y) \ge 0$
$d(x,y) = 0$ iff $x = y$ .
$d(x,y) + d(y,z) \ge d(x,z)$

A metric space is a pair $(X,d)$ where $X$ is a set and $d$ is a metric on $X$ . A ball is a subset of the form $B(x,r) = \{ y \in X : d(x,y) < r\}$ . This set identifies all points of distance smaller than $r$ from $x$ .

Classification of points in the space can be made on the basis of certain properties. Let $E$ be a subset of $X$

a neighbourhood of $x$ is a ball $B(x,r)$ where $r > 0$ .
$x$ is a limit point of $E$ if every neighbourhood of $x$ contains a point of $E$ distinct from $x$ .
$x$ is a point of closure of $E$ if it is a point of $E$ or is a limit point of $E$ .
$x$ is an isolated point of $E$ if it is a point of closure that is not limit point of $E$ .
$x$ is an interior point of $E$ if there is a neighbourhood of $x$ contained in $E$ .

Classification of sets

$E$ is closed if every limit point of $E$ is a point of $E$ .
$E$ is open if every point of $E$ is an interior point of $E$ .
$E$ is perfect if every point of $E$ is a limit point of $E$ and every limit point of $E$ is a point of $E$ .
$E$ is dense in $X$ if every point of $X$ is a point of closure of $E$ .

lemma: neighbourhoods are open sets.

proof: Let $y \in B(x,r)$ . Since $z \in B(y,r-d(x,y))$ implies $d(x,z) \le d(x,y) + d(y,z) < d(x,y) + r - d(x,y) < r$ , it follows that $z \in B(x,r)$ , it follows that $y$ is an interior point of $B(x,r)$ .

lemma: if $p$ is a limit point of $E$ , then every neighbourhood of $p$ contains an infinite number of points of $E$ .

proof: Let $B(x,r)$ be a given neighbourhood of $x$ . Since $x$ is a limit point of $E$ there is a point $e_1 \in B(x,r)$ such that $e_1 \neq x$ . Then $d(x,e_1) > 0$ . Assume distinct points $e_1,e_2,\ldots,e_k$ of $E$ have been selected and such that $d(x,e_i) > d(x,e_{i+1})$ for $i = 1,\ldots,k-1$ . Choose $e_{k+1} \in B(x,d(x,e_k))$ such that $e_{k+1} \in E$ and $d(x,e_{k+1}) > 0$ . Then $\left< e_k \right>$ defines a sequence of distinct points of $E$ distinct from $x$ and has an infinite range.

corollary: a finite set $E$ has no limit points

proof: if it did then any neighbourhood of one of its limit points contains an infinite number of points of $E$ . But $E$ is a finite set and so does not have an infinite number of points.

lemma: a set is open iff its complement is closed.

proof:

Suppose $E$ is an open set. Suppose $E^C$ is not closed. Then there is a limit point $x$ of $E^C$ that is a point of $E$ . But every point of $E$ is an interior point of $E$ , and so there is a neighbourhood of $x$ contained in $E$ . But by the prior lemma, every neighbourhood of a limit point of $E^C$ contains an infinite number of points of $E^C$ , which is false. So $E^C$ is closed.

Conversely, let $E^C$ is closed. Suppose that $E$ is not open. Then there is a point of $x \in E$ that is not an interior point of $E$ . And so for every neighbourhood of $x$ there is a point of $E^C$ contained in it. Since this point is distinct from $x$ , it follows that $x$ is limit point of $E^C$ . But since $E^C$ is closed, it contains all its limit points and so $x \in E^C$ , which is false. Hence $E$ is open.

theorem [set properties of open/closed sets]

let $F$ be a collection of open sets. Then $\cup F$ is an open set
let $F$ be a collection of closed sets. Then $\cap F$ is a closed set
let $F$ be a finite collection of open sets. Then $\cap F$ is an open set
let $F$ be a finite collection of closed sets. Then $\cup F$ is a closed set

proof:

Let $F$ is a collection of open sets. Let $x \in \cup F$ . Then there is an open set $f \in F$ such that $x \in f$ . But since every point of $f$ is an interior point of $f$ there is a neighbourhood $B$ of $x$ contained in $f$ . But since $f$ is a subset of $\cup F$ it follows that $x$ is an interior point of $\cup F$ . Hence every point of $\cup F$ is an interior point of $F$ and so $F$ is open.

Let $F$ be a collection of closed sets. Then $\cap F = (\cup F^C)^C$ is the complement of the union of a collection of open sets, nd therefore is a closed set.

For the latter cases, it is sufficient to consider two sets and appeal to mathematical induction to establish the general property.

Let $a,b$ be open sets. Let $x \in a \cap b$ . Then $x$ is point of $a$ and therefore an interior point of $a$ . Hence there is a ball $B(x,r)$ contained in $a$ . Analgously, there is a ball $B(x,s)$ contained in $b$ . So $B(x,\min(r,s))$ is a ball contained in $a \cap b$ . Hene $x$ is an interior point of the intersection and therefore an open set.

Let $a,b$ be closed sets. Then $a \cup b = (a^C \cap b^C)^C$ is the complement of the intersection of open sets and therefore is a closed set.

lemma: let $cl(E)$ denote the set of points of closure of $E$ . Then

$cl(E)$ is a closed set
$E = cl(E)$ iff $E$ is closed
$cl(E)$ is the smallest closed set containing $E$ .

proof:

Let $x$ be a limit point of $cl(E)$ . Suppose that $x$ is not a point of $cl(E)$ . Then it is not a point of $E$ and it is not a limit point of $E$ . Hence there is a neighbourhood of $x$ that contains no points of $E$ . But since $x$ is a limit point of $cl(E)$ every neighborhood of $x$ contains a point of $cl(E)$ and so contains a point of $E$ . But this is false. Hence $x$ is a point of $cl(E)$ . Since this applies to every limit point, $cl(E)$ is a closed set.

Let $E = cl(E)$ . Then by the prior result, $E$ is a closed set. Conversely, suppose $E$ is a closed set. Then every limit point of $E$ is a point of $E$ , and so $E = cl(E)$ .

Let $F$ be a closed set containing $E$ . Let $x$ be a limit point of $E$ . Then every neighbourhood of $x$ contains a point of $E$ and therefore $F$ . Hence $x$ is a limit point of $F$ . But since $F$ is closed, $x$ is a point of $F$ . Hence $F$ contains all the limit points of $E$ , and so it contains the closure of $E$ . Hence the intersection of all closed sets containing $E$ also contains the closure of $E$ . But the intersection of a collection of closed sets is a closed set and since the closure of $E$ is contained in this collection, it follows that the closure of $E$ is equal to the intersection of all closed sets containing $E$ and so is the smallest such set.

definition: a subspace $Y$ of the metric space $(X,d)$ is the metric space $(Y, d_Y = d | Y\times Y)$ .

lemma: A set $O$ is open in metric space $(Y, d_Y)$ iff there is an set $G$ open in $(X,d)$ such that $O = G \cap Y$ .

proof: Suppose that $O = G \cap Y$ where $G$ is an open set of $X$ .Let $x \in O$ . Then there is $B(x,r) \subseteq G$ . But then $B_Y(x,r) = B(x,r) \cap Y$ is a neighbourhood of $x$ in the space $Y$ , and is such that $B_Y(x,r) \subseteq O$. Hence $O$ is open in $Y$ .

Conversely, suppose that $O$ is an open set of the space $Y$ . Then for each point $x \in O$ there is a $B_Y(x,r(x)) \substeq O$ . Define $G = \cup \{ B(x,r(x)) : x \in O\}$ . Then $G$ is an open set of $X$ and is such that $G \cap Y = \cup \{ B_Y(x,r(x) : x \in O \}$ . But the RHS is both contained in $O$ and contains $O$ and so is identical to $O$ . Hence $G \cap Y = O$ .

Compact Sets

definition: An cover for a set $E$ is a collection of sets whose union contains $E$ . An open cover is a cover comprised of open sets. A subcover is a subset of cover that is a cover.

definition: a subset $K$ of a metric space $X$ is compact if every open cover of $K$ has a finite subcover.

lemma: suppose $K \subseteq Y \subseteq X$ . Then $K$ is compact in space $Y$ iff it is compact in space $X$ .

proof: suppose $K$ is compact in space $X$ . Let $C$ be an open cover of $K$ in space $Y$ . Then each set $O \in C$ has an open set $G(O)$ in space $X$ such that $O = G(O) \cap Y$ . But then $D = \{G(O) : O \in C\}$ is an open cover of $K$ in space $X$ . Since $K$ is compact in this space, it has a finite subcover $\{G(O_1),\ldots,G(O_n)\}$ . But then $\{O_1,\ldots,O_n\}$ is a finite subcover of $C$ . Hence $K$ is compact in space $Y$ .

Conversely, suppose $K$ is compact in space $Y$ . Let $C$ be an open cover of $K$ in space $X$ . Then $C_Y = \{O \cap Y : O \in C\}$ is an open cover of $K$ in space $Y$ and so has a finite subcover. But this implies that $C$ has a finite subcover. Hence $K$ is compact in space $X$ .

lemma: compact sets of metric spaces are closed.

proof: Let $K$ be a compact subset of the metric space $X$ . Let $p$ be a point in $K^C$ . For $q \in K$ , $d(p,q) > 0$ and so there are disjoint neighbourhoods $V_q$ of $q$ and $W_q$ of $p$ . Since $\{V_q : q \in K\}$ is an open cover of $K$ , there is a finite sequence $q_1,\ldots,q_n$ such that $\{V_{q_i} : i = 1,\ldots,n\}$ . is a subcover. But then $\cap_i W_{q_i} \cap V_{q_j} = \emptyset$ and so $\cap_i W_{q_i} \cap \cup_j V_{q_j} = \emptyset$ . Hence $\cap_i W_{q_i} \subseteq K^C$ , and so $x$ is an interior point of $K^C$ . Hence $K^C$ is open, and so its complement $K$ is closed.

lemma: closed subsets of compact sets are compact

proof: Let $F$ be a closed subset of the compact set $K$ . Let $C$ be an open cover of $F$ . Then $F^C$ is open and $D = C \cup \{F^C \}$ is an open cover of $X$ and therefore of $K$ , and so there is a finite subcover of $D$ . But this implies there is a finite subcover of $C$ . Hence $F$ is compact.

corollary: if $F$ is closed and $K$ is compact then $F \cap K$ is compact.

proof: since $F \cap K$ is a closed subset of the compact set $K$ , it follows that it is compact.

theorem: If $\mathcal{K}$ is a collection of compact sets such that every finite subcollection has a nonempty intersection, $\cap \mathcal{K}$ is nonempty.

proof: Suppose that $\cap \mathcal{K}$ is empty. Then $\cup \mathcal{K}^C = X$ with the LHS being the union of a collection of complements of compact sets and therefore a collection of open sets. Choose any $k \in \mathcal{K}$ . Then since $k$ is compact, there is a finite subcover of $\mathcal{K}^C$ . Hence $k \subseteq k_1^C \cup \ldots k_n^C$. But this implies that $k \cap k_1 \cap \ldots k_n = \emptyset$, which is false. So the supposition is false.

corollary: If $\left< K_n \right>$ is a non-increasing sequence of nonempty compact sets, then $\cap_n K_n$ is noempty.

proof: the intersection of every finite subcollection contains the compact sets with the largest index and since every set in the collection is nonempty, the intersection is nonempty. So the intersection is nonempty.

lemma: If $E$ is an infinite subset of compact set $K$ then $K$ contains a limit point of $E$ .

proof: suppose $K$ contains no limit point of $E$ . Then every point of $K$ has a neighbourhood that contains at most one point of $E$ . This is an open cover of $K$ , and so has a finite subcover. But this implies $E$ has a finite subcover. But a finite subcover of this collection only covers a finite number of points of $E$ , which is a contradiction. So the initial supposition is false.

Metric Spaces and Compactness

Metric

Compact Sets

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